Single line diagram

2.  Representation of essentials of a system in a most
simplified form.
 Main components of a power plant
 Generation plant
 Transmission line
 Distribution line
 Industrial, commercial and residential load
 Nowadays all of these are in the form of three phase

3.  A one-line diagram or single-line diagram (SLD) is a
simplified notation for representing a three-phase power
system.
 The one-line diagram has its largest application in power
flow studies.
 Electrical elements such as circuit breakers, transformers,
capacitors, bus bars, and conductors are shown by
standardized schematic symbols.
 Instead of representing each of three phases with a separate
line or terminal, only one conductor is represented.
 It is a form of block diagram graphically depicting the paths
for power flow between entities of the system.
 Elements on the diagram do not represent the physical size
or location of the electrical equipment, but it is a common
convention to organize the diagram with the same left-to-
right, top-to-bottom sequence as the switchgear or other
apparatus represented.

5. Impedance diagram and reactance diagram
 Synchronous generator are replaced by:
 Transformer is replaced by:
 Transmission lines are replaced by:
 In many power system studies synchronous generator
resistance, resistance of transformer windings,
resistance of transmission lines, line charging and
magnetizing circuits of transformer are neglected.
 The impedance diagram then become reactance
diagram

6. Example 8.2
 A three phase synchronous generator delivers 10 MVA at a
voltage of 10.5 kV. The line impedance is 5 ohm. Determine the
voltage drop in the line in per unit and in volts. Use the reference
base as 10 MVA at 11 kV.
 S3φ= 10 MVA
 (Sb)3φ = 12MVA
 Vl=10.5kV
 Vlb= 11kV
 Vpu=(Vl)pu= Vl/Vlb=10.5/11
 Spu=S3φ/(Sb)3φ=10/12
 Ipu=Spu/Vpu=10/12*11/10.5= 0.8730158
 Zb= Vlb
2/ (Sb)3φ =112 /12
 Zpu= ZΩ/Zb=5*12/ 112 =0.4958677

7.  Voltage drop in line per phase:
 ΔVpu= Zpu* Ipu= 0.4958677 *0.8730158=0.4329pu
 ΔVpu=Vactual/Vb
 ΔVactual= ΔVpu*Vb=0.4329*11kV/root(3)=2749.36 V